Standing waves

By Michael Lasaleta 39607122

String vibrations

• A young physicist student at UBC by the name Kanye, fell in love with a woman named Taylor that devoted her life to music. Everyday the young boy would watch admiringly from afar. The woman never stopped playing the guitar and the boy didn’t want to interrupt her playing as she was dedicated to becoming a professional guitar player.

String vibrations

• One day he interrupted her playing and told her “I’m going to let you finish, but I want you to date me at least one time.” She smiled politely, and said she would only go on a date with him if he could impress her using her guitar. The guitar was something special to her and she knew every fact about it. The physics student was scared he would not be able to impress her since he had no musical talent. He paused, and quickly examined the guitar.

• He told the woman that her guitar had a string that was 0.65m long and has a linear mass density of 4.0x10-

4kg/m. The tension in the string is kept at 80.0N. The young girl nodded with excitement because she knew he was correct but was not satisfied.

Problem

• She confirmed that the guitar had a string that was 0.65m long and has a linear mass density of 4.0x10-4kg/m. The tension in the string is kept at 80.0N. She asked the following questions to the young man when she plucked the string.

• What is the wave speed in the string?

• If the string is plucked and allowed to vibrate, approximately how many times would a wave reflect from one end of the string in one second?

• What are the wavelengths and frequencies of the first three normal modes of vibration of the string? What is the spacing between two consecutive nodes for each mode?

• The young man was told by the woman that the pitch was too high and that he should decrease the fundamental frequency by 10 hz to capture her heart. By what amount should the tension be changed for the young man to capture the woman’s heart?

Known quantities

• Length of string = 0.65m

• u = 4.0x10-4kg/m

• T = 80.0N

Solution

• What is the wave speed in the string?

• The boy explained that the wave speed in the string is related to tension and the linear mass density of the guitar string.

• v=√(80.0N4.00x10-4kg/m)

• =447.21m/s

Solution• If the string is plucked and allowed to vibrate, approximately how many times

would a wave reflect from one end of the string in one second?

• The time that the waves take to travel from one end to the other is related to string length and wave speed:

• Δ t= (string length/wave speed)

• Δ t= (0.65m)/(447.21m/s)

• Δ t= 1.45x10-3s

• Each reflection requires 2 trips, one forward and one backwards. Therefore, we can divide 1 second by two times Δ t.

• 1.0s/(2xΔ t) = 1.0s/(2x1.45x10-3s) = 344

Solution• What are the wavelengths and frequencies of the first two normal modes of vibration

of the string? What is the spacing between two consecutive nodes for both mode?

For the first harmonic:

λ1= 2L = 2x0.65m = 1.3m

f1= v/λ1 = (447.21m/s)/1.3m = 334Hz

The distance between consecutive nodes = λ1/2 = 1.3m/2 = 0.65m

For the second harmonic:

λ2=2L/2 = 2(0.65m)/2=0.65m

f2= 2xf1= 2x344Hz = 688Hz

The distance between consecutive nodes = λ2/2 = 0.65m/2 = 0.325m

Solution• The young man was told by the woman that the pitch was too high and

that he should decrease the fundamental frequency by 10 hz to capture her heart. By what amount should the tension be changed for the young man to capture the woman’s heart?

New fundamental frequency,

f1’= 334Hz - 10Hz= 324Hz

Since fundamental frequency needs to decrease, tension must be decreased.

f1’= (1/(2L)(√(T’/u)

T’=4L2u(f1’)2

Fractional decrease in tension need to lower the frequency by 10.0Hz is,

(T-T’)/T= [(f1)2-(f1’)2]/(f1)2 = (344Hz)2-(324Hz)2/(344Hz)2 = 0.1129

Therefore, the tension in the string needs to be decreased by 0.1129x80.0N =9.03N

Result

• The woman was so amazed by his knowledge about guitars that she fell in love with the physicist and they got married.

THANK YOU!-Michael Lasaleta